Optimal. Leaf size=235 \[ -\frac {\left (1-c^2 x^2\right )^3}{b c x (a+b \text {ArcSin}(c x))}-\frac {25 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2}-\frac {25 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2}-\frac {25 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2}-\frac {25 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2}-\frac {\text {Int}\left (\frac {\left (1-c^2 x^2\right )^2}{x^2 (a+b \text {ArcSin}(c x))},x\right )}{b c} \]
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Rubi [A]
time = 0.34, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \text {ArcSin}(c x))^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}-\frac {(5 c) \int \frac {\left (1-c^2 x^2\right )^2}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac {5 \text {Subst}\left (\int \frac {\cos ^5(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac {5 \text {Subst}\left (\int \left (\frac {5 \cos (x)}{8 (a+b x)}+\frac {5 \cos (3 x)}{16 (a+b x)}+\frac {\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac {5 \text {Subst}\left (\int \frac {\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac {25 \text {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac {25 \text {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}-\frac {\left (25 \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b}-\frac {\left (25 \cos \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac {\left (5 \cos \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac {\left (25 \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b}-\frac {\left (25 \sin \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac {\left (5 \sin \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac {25 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b^2}-\frac {25 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2}-\frac {25 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b^2}-\frac {25 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2}-\frac {\int \frac {\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ \end {align*}
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Mathematica [A]
time = 8.90, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x (a+b \text {ArcSin}(c x))^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [A]
time = 0.86, size = 0, normalized size = 0.00 \[\int \frac {\left (-c^{2} x^{2}+1\right )^{\frac {5}{2}}}{x \left (a +b \arcsin \left (c x \right )\right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{x \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [A]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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